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Ca + 2HCl = CaCl₂ + H₂
c=4.50 mol/l
v=2.20 l
n(HCl)=cv
m(Ca)/M(Ca)=n(HCl)/2
m(Ca)=M(Ca)cv/2
m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g
198 grams of Ca are needed
Answer:
The elements in group 13 and group 15 form a cation with a -3 charge each.
9ml will be given for the case of dosage calculation order: 3 mg available: 2 mg per 6 ml
Conversion factors are necessary for dosage calculation, such as when translating from pounds to kilograms or liters to milliliters. This approach, which is straightforward in design, enables physicians to deal with different units of measurement and convert factors to arrive at the solution.
dosage calculation techniques serve as a second or third check on the accuracy of the previous computation techniques. Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method are the three main approaches for dosage calculation. dosage calculations are frequently prescribed and labeled based on their weight or, for solutions, their strength, which is the amount of weight dissolved or suspended in a given volume.
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<h3><u>Answer;</u></h3>
Empirical formula = C₂H₃O
Molecular formula = C₁₄H₂₁O₇
<h3><u>Explanation</u>;</h3>
Empirical formula
Moles of;
Carbon = 55.8 /12 = 4.65 moles
Hydrogen = 7.04/ 1 = 7.04 moles
Oxygen = 37.16/ 16 = 2.3225 moles
We then get the mole ratio;
4.65/2.3225 = 2.0
7.04/2.3225 = 3.0
2.3225/2.3225 = 1.0
Therefore;
The empirical formula = <u>C₂H₃O</u>
Molecular formula;
(C2H3O)n = 301.35 g
(12 ×2 + 3× 1 + 16×1)n = 301.35
43n = 301.35
n = 7
Therefore;
Molecular formula = (C2H3O)7
<u> = C₁₄H₂₁O₇</u>
